f(x)=∫[0,+∞) f(x,y)dy
=∫[0,+∞) e^(-x-y)dy
=-e^(-x-y)[0,+∞)
=e^(-x)
同理
f(y)=∫[0,+∞) f(x,y)dx
=∫[0,+∞) e^(-x-y)dx
=-e^(-x-y)[0,+∞)
=e^(-y)
f(x)*f(y)=f(x,y)
因此x,y独立
P(X<1,Y<1)
=∫[0,1]∫[0,1] f(x,y)dydx
=∫[0,1] f(x)dx*∫[0,1] f(y)dy
=e^(-x)[0,1]*e^(-y)[0,1]
=(1-1/e)^2